Chapter 4 Homework

B. Although the gravitational constant was unknown in Newton's time, it was still possible to compare the mass of the Earth to that of the Sun. Explain.
The short answer is that forces between earth, sun, and moon can all be computed, but only in arbitrary units (because G was unknown). Thus the relative magnitudes of the forces and hence masses could be computed. The details - for those of you who want to slog through a bit of algebra - follow:
Applying the law of universal gravitation and Newton's 2nd law to the moon's orbit,
Fearth on moon = GMearthMmoon/R2earth-moon = Mmoon *amoon
Note that the mass of the moon is a common factor on both sides of the equation and so cancels out, leaving
GMearth/R2earth on moon = amoon
Since Newton knew the size of moon's orbit and its timing, he could calculate its speed (1012m/s) and acceleration (.0027m/s2). The only thing that is not known is the product GMearth, so solving for this one gets GMearth = amoon*R2earth on moon = 3.9E14.
The same calculatons can be done for the earth-sun system (R=1.5E11m, v=29,900m/s, and a=.0060m/s2. Computing GMsun = aearth*R2sun-earth = 1.3E20. The ratio (GMsun) / (GMearth) = Msun / Mearth = 340,000.

C. One of Galileo's arguments in favor of the Copernican theory was that Venus had phases like the Moon. Explain how this is relevant. Could Tycho's scheme also explain this observation? Could Ptolemy's scheme explain this observation?
Ptolemy's system would have a very difficult time explaining the phases of Venus - becuase these require very specific relative orientations between the earth, sun, and venus. Both of the models of Copernicus and Tycho Brahe (circular orbits with Venus orbiting sun - Copernicus had Earth orbiting sun while Brahe had Sun orbiting earth)

3. Calculate the force between two objects of mass 10kg and 100kg, separated by 0.2m.
F = GMM/r2 = (6.67E-11 Nm2/kg2) *10kg *100kg / (0.2m)2 = 6.67E-8/.04 N = 1.7E-6 N
This is roughly the weight of .2 milligrams.